Let $g(x)=\tan(x)$. Find $g'\left(\dfrac{\pi}{4}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{2}$ (Choice B) B $1$ (Choice C) C $\dfrac{\sqrt{2}}{2}$ (Choice D) D $2$
Answer: Let's first find $g'(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{4}$. Recall that the derivative of $\tan(x)$ is $\dfrac{1}{\cos^2(x)}$, or $\sec^2(x)$. [Is there a way to know this without memorizing?] So $g'(x)=\dfrac{1}{\cos^2(x)}$. Now let's plug $x={\dfrac{\pi}{4}}$ into $g'$ : $\begin{aligned} &\phantom{=}g'\left({\dfrac{\pi}{4}}\right) \\\\ &=\dfrac{1}{\cos^2\left({\dfrac{\pi}{4}}\right)} \\\\ &=\dfrac{1}{\left(\dfrac{\sqrt 2}{2}\right)^2} \\\\ &=\dfrac{1}{\left(\dfrac{2}{4}\right)} \\\\ &=2 \end{aligned}$ In conclusion, $g'\left(\dfrac{\pi}{4}\right)=2$.